Mean Power in a Resistive Load | CAIE 9702 Paper 4 Notes

Paper 4

Physics9702

Mean Power in a Resistive Load

Mean power can be calculated with r.m.s. values. For a sine wave, it is also half the peak power.

Keep peak values and r.m.s. values in separate methods

\langle P\rangle=I_{\mathrm{rms}}^2R

Symbols

\langle P\rangle=\dfrac{V_{\mathrm{rms}}^2}{R}

\langle P\rangle=\dfrac{1}{2}P_0\quad\text{for a sine wave}

A peak current must be used to find peak power. An r.m.s. current must be used to find mean power.

Worked example

Peak power and mean power

A sinusoidal current has peak value 3.5 A and flows through a 680 Ω resistor. Find its peak and mean powers.

Use the peak current for peak power.
$P_0=3.5^2\times680$
$P_0=8.33\times10^3\ \mathrm{W}$
Mean power is half the peak power.
$\langle P\rangle=\dfrac{1}{2}P_0$
$\langle P\rangle=4.17\times10^3\ \mathrm{W}$

Answer

P_0=8.33\ \mathrm{kW}

\langle P\rangle=4.17\ \mathrm{kW}

This exact comparison was required in 2024 (9702/42/O/N/24 Q8(c)).

Common mistake

Putting a peak current into the r.m.s. power equation makes the mean power twice too large.

Your turnpower check

A 12 V r.m.s. sinusoidal voltage is connected across a 24 Ω resistor. Calculate the mean power and the peak power.

Show worked answer

\langle P\rangle=\dfrac{12^2}{24}=6.0\ \mathrm{W}

Answer

P_0=2\langle P\rangle=12\ \mathrm{W}

Mean Power in a Resistive Load

Mean power can be calculated with r.m.s. values. For a sine wave, it is also half the peak power.

\langle P\rangle=I_{\mathrm{rms}}^2R

Symbols

\langle P\rangle=\dfrac{V_{\mathrm{rms}}^2}{R}

\langle P\rangle=\dfrac{1}{2}P_0\quad\text{for a sine wave}

A peak current must be used to find peak power. An r.m.s. current must be used to find mean power.

Worked example

Peak power and mean power

A sinusoidal current has peak value 3.5 A and flows through a 680 Ω resistor. Find its peak and mean powers.

Use the peak current for peak power.
$P_0=3.5^2\times680$
$P_0=8.33\times10^3\ \mathrm{W}$
Mean power is half the peak power.
$\langle P\rangle=\dfrac{1}{2}P_0$
$\langle P\rangle=4.17\times10^3\ \mathrm{W}$

Answer

P_0=8.33\ \mathrm{kW}

\langle P\rangle=4.17\ \mathrm{kW}

This exact comparison was required in 2024 (9702/42/O/N/24 Q8(c)).

Common mistake

Putting a peak current into the r.m.s. power equation makes the mean power twice too large.

Your turnpower check

A 12 V r.m.s. sinusoidal voltage is connected across a 24 Ω resistor. Calculate the mean power and the peak power.

Show worked answer

\langle P\rangle=\dfrac{12^2}{24}=6.0\ \mathrm{W}

Answer

P_0=2\langle P\rangle=12\ \mathrm{W}